1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
stiv31 [10]
3 years ago
9

Object a has a mass of 20 g

Physics
1 answer:
Anarel [89]3 years ago
7 0
And.. where is the rest of the question?
You might be interested in
Find the area under the standard normal curve to the right of z=0.49. round your answer to four decimal places, if necessary.
VikaD [51]

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

<h3>What is  the standard normal curve?</h3>

The horizontal axis is approached by the standard normal bend as it extends indefinitely both in directions without ever being touched by it. The center of the bell-shaped, z=0 standard normal curve. Between z=3 and z=3, almost the entire area underneath the standard normal curve is located.

<h3>Use of the standard normal curve:</h3>

Use the normal distribution's standard form to calculate probability. Since the standard normal distribution is indeed a probability distribution, the probability that a variable will take on a range of values is indicated by area of the curve between two points. 100% or 1 is the total area beneath the curve.

<h3>According to the given data:</h3>

the region to the left of the standard normal curve,

z=0.49

To the right of,

z = 2.05

So,

The area will be:

= P[z < 0.49] + P[ z >2.05]

= P[z < 0.49] + 1 -  P[ z < 2.05]

= .6879 + 1 - .9798

= 0.7081

Left of z = 0.49 and right of z = 2.05, the area underneath the standard normal curve is equal to 0.7081.

To know more about standard normal curve visit:

brainly.com/question/12972781

#SPJ4

I understand that the question you are looking for is:

Find the area under the standard normal curve to the left of z = 0.49 and to the right of z = 2.05. Round your answer to four decimal places, if necessary.

4 0
1 year ago
A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude
Darya [45]

Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

\omega = \sqrt{\frac{k}{m} } \\\\\omega =  \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

8 0
2 years ago
A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an
Nat2105 [25]

Answer:

a)T=0.0031416s

b)v_{max}=6.283\frac{m}{s}

c) E=0.1974J

d)F=80N

e)F=40N

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by X=Acos(\omega t +\phi)    (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)   (2)

For the acceleration

\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)   (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

A\omega^2=8x10^{3}\frac{m}{s^2}

Since we know the amplitude A=0.002m  we can solve for \omega like this:

\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}

And we with this value we can find the period with the following formula

T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens cos(\omega t +\phi)=1, and we also know that the maximum acceleration is given by::

|\frac{d^2X}{dt^2}|=A\omega^2

So then we have that:

F=ma=mA\omega^2

And since we have everything we can find the force

F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N

6) Part e

When the mass it's at the half of it's maximum displacement the term cos(\omega t +\phi)=1/2 and on this case the acceleration would be given by;

|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}

And the force would be given by:

F=ma=\frac{1}{2}mA\omega^2

And replacing we have:

F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N

8 0
3 years ago
What energy transformation takes place in the<br> solar panel?
aliina [53]
I think it’s the first option
3 0
3 years ago
A spring-mass system has a spring constant of 3 Nm. A mass of 2 kg is attached to the spring, and the motion takes place in a vi
frosja888 [35]

Answer:

The answer to the question

The steady state response is u₂(t) = -\frac{3\sqrt{2} }{2}cos(3t + π/4)

of the form R·cos(ωt−δ) with R = -\frac{3\sqrt{2} }{2}, ω = 3 and δ = -π/4

Explanation:

To solve the question we note that the equation of motion is given by

m·u'' + γ·u' + k·u = F(t) where

m = mass = 2.00 kg

γ = Damping coefficient = 1

k = Spring constant = 3 N·m

F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)

Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)

The homogeneous equation 2·u'' + u' + 3·u is first solved as follows

2·u'' + u' + 3·u = 0 where putting the characteristic equation as

2·X² + X + 3 = 0 we have the solution given by \frac{-1+/-\sqrt{23} }{4} \sqrt{-1} =\frac{-1+/-\sqrt{23} }{4} i

This gives the general solution of the homogeneous equation as

u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t)

For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)

Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which  2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t)  we have

3·B-15·A = 27

3·A +15·B = 18

Solving the above linear system of equations we have

A = -1.5, B = 1.5 and  u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)

u₂(t) = 1.5·(sin(3·t) - cos(3·t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4)

The general solution is then  u(t) = u₁(t) + u₂(t)

however since u₁(t) = e^{(-1/4t)} (C_1cos(\frac{\sqrt{23} }{4}t) + C_2sin(\frac{\sqrt{23} }{4}t) ⇒ 0 as t → ∞ the steady state response = u₂(t) = -\frac{3\sqrt{2} }{2}·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where

R = -\frac{3\sqrt{2} }{2}

ω = 3 and

δ = -π/4

8 0
3 years ago
Other questions:
  • Why does the sound of something moving away from you seem to change to lower and lower pitch
    10·1 answer
  • Defined the term electromagnetic induction and explain how a generator uses electromagnetic induction​
    10·1 answer
  • A woman is standing at the rim of a nonuniform cylindrical horizontal platform initially at rest. The platform is free to rotate
    15·1 answer
  • Two runners start a race. After 2 seconds, they both have the same velocity. If they both started at the same time, how do their
    10·2 answers
  • Suppose you have two point charges of opposite sign. As you move them farther and farther apart, the potential energy of this sy
    7·2 answers
  • An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piec
    15·1 answer
  • A kid applies a force of 70 N to a ball over 1.1 meters. How much work did he do on the ball?
    14·1 answer
  • 1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
    10·1 answer
  • What is the smallest particle in the Universe?
    10·2 answers
  • What time will the lunar eclipse happen central time?.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!